The universe as a sphere.

Consider a uniform sphere that has mass M and radius R. The gravitational energy of an isotropic sphere is given by \displaystyle -\frac{16}{15} \pi^2 G \sigma^2 R^5 so using \displaystyle \sigma = \frac{3M}{4 \pi R^3} we obtain

\displaystyle U = -\frac{3 G M^2}{5 R}

The energy is given by \displaystyle \int \frac{dm c^2}{\sqrt{1 - v^2/c^2}} and defining \displaystyle v = \frac{r}{R} \dot{R} we obtain

\displaystyle T = \int \frac{dm c^2}{\sqrt{ \displaystyle 1 - \dot\zeta^2 \left( \frac{r}{R} \right)^2}}

where \displaystyle \dot\zeta = \frac{\dot{R}}{c}. So

\displaystyle T = 4 \pi \sigma R^3 c^2 \int_0^R d(r/R) \frac{(r/R)^2}{\sqrt{1 - \dot\zeta^2 (r/R)^2}} = 4 \pi \sigma R^3 c^2 \int_0^1 dx \frac{x^2}{\sqrt{1 - \dot\zeta^2 x^2}}

Using \displaystyle \sigma = \frac{3M}{4 \pi R^3} we can write

\displaystyle T = 3Mc^2 \frac{\sin^{-1}(\dot\zeta) - \dot\zeta \sqrt{1 - \dot\zeta^2}}{2 \dot\zeta^3} = \frac{3Mc^2}{f(\dot\zeta)}.

Where \displaystyle f(z) = \frac{2z^2}{\sin^{-1}(z) - z \sqrt{1-z^2}} and we focus on the function f(z), the plot is given by

The function f(z) looks like a circle. Note that f(0)=3 and f(1) = 4/\pi – so we consider \displaystyle \frac{f(z) - 4/\pi}{3-4/\pi} – the plot is given by

We therefore propose \displaystyle \frac{f(z)-4/\pi}{3-4/\pi} \simeq \sqrt{1-z^2}.

The total energy is given by

\displaystyle E = T + V = 3Mc^2 \left\{ \frac{1}{f(\dot\zeta)} - \frac{GM}{5Rc^2}\right\}

We assume that E=0 and that c is constant in time, then we obtain

\displaystyle f(\dot\zeta) = \kappa\zeta

where \displaystyle \kappa = \frac{5c^3}{GM} So

\displaystyle \frac{f(\dot\zeta)-4/\pi}{3-4/\pi} = \frac{\kappa\zeta-4/\pi}{3-4/\pi} \simeq \sqrt{1-\dot\zeta^2}

The solution is given by

\displaystyle \zeta = \frac{GM}{c^3} \frac{4}{5\pi} \left\{ 1 + \left(3\pi/4-1\right) \sin\left(\frac{c^3}{GM} \frac{5}{3-4/\pi} t\right) \right\}

Or using R = c\zeta and \displaystyle R_\textit{max} = \frac{3}{5} \frac{GM}{c^2} we can write

\displaystyle R = R_\textit{max} \left\{ 4/3\pi + \left(1 - 4/3\pi\right) \cos\left(\frac{1}{1-4/3\pi} \frac{ct}{R_\textit{max}}\right) \right\}

Then R_\textit{max} = R(0) and R(T_\textit{max}/2)=0. So we obtain

\left[\begin{array}{rcl}  R_\textit{max} &=& \displaystyle \frac{3}{5} \frac{GM}{c^2}\\\\  T_\textit{max} &=& \displaystyle \tau \frac{R_\textit{max}}{c}\\\\  \tau &=& 2(1-4/3\pi) \cos^{-1} \left(1/(3\pi/4-1)\right) \simeq 2.7627547843822696042358013453991...  \end{array}\right.

A plot of the radius of the universe in time:

And a plot of the speed of the radius of the universe in time:

The radius is accelerating at the beginning. Is it possible that we are just at the beginning of the universe so that the expansion is still accelerating?

Here we have just used special relativity. The total energy of the universe in this model simply is zero (kinetic energy is positive – but the gravitational energy is negative) – so before the ‘big bang’ there was simply nothing – not energy – no matter and no time.

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This entry was posted in Cosmology, Mathematics, Physics, Relativity, Universe. Bookmark the permalink.

One Response to The universe as a sphere.

  1. facebook says:

    i love it

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